Physics II Chapter I Solutions

Chapter I - Problems Problem I.1 The voltage from an AC source varies with time as VHtL = Vmax cosHw t + fL where Vmax = 20 V, w = 80 s-1 and f = 0.6 . (a) What is the root-mean-squared voltage, Vrms ? (b) What is the frequency, f ? (c) What is the time closest to t = 0 when the voltage takes on its maximum value. Solution to I.1 (a) Vrms = (b) f = 1 2 Vmax = 20 2 = 14.1 V w 80 = = 12.7 Hz 2p 2p (c) t0 = - f w =- 0.6 = -0, 0075 s 80 Problem I.2 For a sawtooth pattern voltage show that Vrms = 1 3 Vmax . V Vmax t T -Vmax 2 Chapter I - Problems Solution to I.2 V Vmax -Tê2 t Tê2 -Vmax To find the average integrate over one period of the function. One period of a sawtooth voltage may be written as V HtL = Vrms = V2 = 1 Tê2 V2 „ t = Ÿ T -Tê2 1 2 Vmax 2 Tê2 2 K O Ÿ-Tê2 t „ t = T T ï Vrms = Vmax Tê2 t. 1 2 Vmax 2 T 3 K O T T 12 Vmax 3 Problem I.3 An AC voltage source has an rms voltage of 20 V and a frequency of 40 Hz. Suppose the manufacturer's specs say that this source cannot safely put out a larger instantaneous current than 80 mA. (a) What resistance values can be connected across this source? (Just the resistor is connected across the source.) (b) What capacitance values can be connected across this source? (c) What inductance values can be connected across this source? Solution to I.3 The peak current Imax can be no larger 80 mA, thus the largest value of Irms can be Irms = 1 2 Imax = 0.05657 = 0.0566 A. This with Vrms gives the smallest safe impedance. Zmin = Vrms Irms = 20 = 353.55 = 353 W 0.05657 The impedance for just a resistor is R, for just a capacitor is XC and for an inductor is XL . (a) R = Z ¥ Zmin = 353 W (b) Z = Xc = 1 1 1 ¥ Zmin ï C § = = 11.3 mF 2p f C 2 p f Zmin 2 p 40µ353.55 Z 353.55 (c) Z = XL = 2 p f L ¥ Zmin ï L ¥ min = = 1.41 H 2p f Problem I.4 2 p 40 Chapter I - Problems 3 Problem I.4 What is the maximum charge on the plates of a capacitor with capacitance C when connected across an AC source with frequency f and rms voltage Vrms ? Solution to I.4 Qmax = C Vmax = C Vrms 2 Problem I.5 A voltage source of produces an output which varies as a function of time as VHtL = 50 V cosAI300 s-1 M tE. (a) What are the root-mean-squared voltage and frequency of the source? (b) Suppose this voltage is connected across a series combination of a 100 mF capacitor and a 40 W resistor. What is the rms current through the circuit and what is the phase angle? (c) Write the current as a function of time. Solution to I.5 (a) From the function of time we can read off that. Vmax = 50 V and w = 300 s-1 . We can then find Vrms and f from this. Vrms = 1 2 Vmax = 35.344 = 35.4 V and f = w = 47.7 Hz 2p (b) We have R = 40 W and C = 100 µ 10-6 F. Since there is no inductor we have XL = 0. The capacitive reactance is XC = 1 = 33.333 . wC We can then find the impedance and rms current, and then the phase angle. Z= R2 + IXL - XC M2 = 52.068 ï Irms = f = tan-1 XL -XC R Vrms Z = 0.67902 = 0.679 A = -39.8 = -0.695 rad (c) To get the current as a function of time we must first find Imax . Starting from Vmax we can use Imax = Vmax ê Z but since we have already found Irms it is easier to use Imax = 2 Irms = 0.960 A. We saw in the notes that IHtL = Imax cosHw tL ï VHtL = Vmax cosHw t + fL. 4 Chapter I - Problems IHtL = Imax cosHw tL ï VHtL = Vmax cosHw t + fL. When the voltage is varies as cos w t then the current must shift in the reverse direction. VHtL = Vmax cosHw tL ï IHtL = Imax cosHw t - fL = 0.960 A cosAI300 s-1 + 0.695ME Problem I.6 a 30 W 90 mH 50 mF b c d A 50 mC capacitor, a 90 mH inductor and a 30 W resistor are connected in series across a standard outlet in the order listed, C then L then R. An AC multimeter measures the rms voltage and rms current. (a) When the multimeter is set as an ammeter and put into the circuit, what does it read? (b) What does the multimeter set as a voltmeter read when connected between a and b, between b and c, between c and d, between a and c, and between b and d? (c) By what phase angle is the voltage ahead of the current? (d) What is the average power dissipated in the capacitor, in the inductor and in the resistor? Solution to I.6 For a standard outlet Vrms = 120 V and f = 60 Hz. The components have values: C = 50 µ 10-6 F, L = 0.090 H and R = 30 W. Before doing other calculations we must first find the reactances and impedance. XC = 1 = 53.052 = 53.0 W, 2p f C XL = 2 p f L = 33.929 = 33.9 W and Z= R2 + IXL - XC M2 = 35.576 = 35.6 W (a) The rms current is found by Vrms = Irms Z. Irms = Vrms Z = 3.3730 = 3.37 A (b) To get the voltage between two points use Vrms = Irms Z £ where Z £ is the impedance between the two positions. All voltages given here are rms values. The impedance between a and b is the capacitive reactance. Zab = XC ï Vab = Irms XC = 179 V The impedance between b and c is the inductive reactance. Zbc = XL ï Vbc = Irms XL = 114 V The impedance between c and d is the resistance. Zcd = R ï Vcd = Irms R = 101 V Between a and c we have XC and XL but no resistance. Zac = IXL - XC M2 = 19.122 ï Vac = Irms Zac = 64.5 V Chapter I - Problems 5 IXL - XC M2 = 19.122 ï Vac = Irms Zac = 64.5 V Zac = Between b and d we have XL and R but no capacitor, so XC = 0. Zbd = (c) The phase angle is found by tan f = R2 + XL 2 = 45.2901 ï Vbd = Irms Zbd = 153 V XL -XC R . f = tan-1 XL -XC R = -32.5 ° 2 R. All the (d) The average power dissipated in an AC circuit is  = Vrms Irms cos f. For a series RCL circuit this reduces to  = Irms power dissipated in an AC circuit is lost in the resistor. 2 R = 341 W and  = 0 =  R = Irms C L Problem I.7 The resonance circuit of a radio tuner is uses a 1.4 mH inductor and a variable capacitor to tune a 99.7 MHz radio station. What value of the capacitance is needed. Solution to I.7 f= 1 2p LC 1 ï 99.7 µ 106 = 2p ï C = 1.82 pF 1.4µ10-6 µC Problem I.8 A 2 W resistor, an 8 mH inductor and a 50 mF capacitor are connected in series. (a) What is the resonant frequency? (b) Suppose this is connected across a voltage source with a fixed Vrms = 5 V and a variable frequency. At what frequency is the rms current its maximum and what is the maximum value of Irms ? Solution to I.8 (a) The resonance frequency is fres = 1 2p 1 = LC 2p = 252 Hz 0.008µ50µ10-6 (b) The impedance is its smallest at resonance, Irms = Vrms ê Z, so the maximum current is at the resonance frequency. Z = Zmin = R ï maxH Irms L = Vrms Zmin = Vrms R = 5 = 2.5 A 2 6 Chapter I - Problems Problem I.9 An ideal transformer is used to raise the voltage from standard household voltage to an rms voltage of 240 V. If the primary coil has 300 turns carrying an rms current of 5 A, then how many turns are in the secondary coil and what is the current through the secondary coil? What is the power delivered to the transformer? Solution to I.9 V2 N I N 240 5 = 2= 1 ï = 2 = ï N2 = 600 and I2 = 2.5 A 120 I2 V1 N1 I1 300  = V1 I1 = 120 µ 5 = 600 W